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Calculating coefficient of static friction without a mass.

  1. Mar 30, 2011 #1
    1. The problem statement, all variables and given/known data
    A car is travelling at 30m/s around a bend of radius 70m. What coefficient of static friction between the tires and the road is required to ensure the car will not slip?


    2. Relevant equations



    3. The attempt at a solution
    I know that the centripetal acceleration is a=30^2/70
    and I know that this acceleration has to be canceled out by the frictional force for the car to not slip, but I am not sure out to calculate frictional force F=mA without a mass for the car.
     
  2. jcsd
  3. Mar 30, 2011 #2

    Doc Al

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    Staff: Mentor

    Maybe you don't need the actual mass. Try calling the mass 'm' and see what happens.

    (Tip: Solve things symbolically as much as possible. Only use the calculator at the end.)
     
  4. Mar 30, 2011 #3
    I'm guessing that since mass will not change and the two forces have to be equal the m will cancel out but I still do not know how to calculate the acceleration into newtons for this situation.
     
  5. Mar 30, 2011 #4

    Doc Al

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    Staff: Mentor

    You're right. But no need to guess: Try it and see!
    I assume that you mean you cannot calculate the force in Newtons. True, without the mass, you cannot calculate the force. So what? You can figure out the coefficient of friction, which is all you need.
     
  6. Mar 30, 2011 #5
    I'm sorry but I am totally lost now..
    What I have done so far: F(r)=mv^2/r=F(fr)
    F(r) being centripetal force and F(fr) being frictional force.
    But I do not know the equation for calculating coefficient of static friction.

    Thanks for helping me out.
     
  7. Mar 30, 2011 #6

    Doc Al

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    How does the maximum static friction force relate to the normal force between surfaces?
     
  8. Mar 30, 2011 #7
    Oh I was completely ignoring normal force!
    F=u(9.8m/s^-2)m
    mv^2/r=u(9.8m/s^-2)m
    (30)^2/70=u(9.8)

    Right?
     
  9. Mar 30, 2011 #8

    Doc Al

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    Right!

    (This assumes that the road is flat--no banking angle.)
     
  10. Mar 30, 2011 #9
    You have been the best help.
    Thank you for not just telling me the answer cause now I have actually learnt how to solve these for myself. :smile:

    Thank you!
     
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